确定系数A,B 使下式成立.∫(dx/(a+bcosx)^2) = Asinx/(a+bcosx) + B∫(dx/(a+bcosx))

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确定系数A,B 使下式成立.∫(dx/(a+bcosx)^2) = Asinx/(a+bcosx) + B∫(dx/(a+bcosx))

确定系数A,B 使下式成立.∫(dx/(a+bcosx)^2) = Asinx/(a+bcosx) + B∫(dx/(a+bcosx))
确定系数A,B 使下式成立.
∫(dx/(a+bcosx)^2) = Asinx/(a+bcosx) + B∫(dx/(a+bcosx))

确定系数A,B 使下式成立.∫(dx/(a+bcosx)^2) = Asinx/(a+bcosx) + B∫(dx/(a+bcosx))
∫ (1/(a + b Cos[x])^2) dx
=-(2 a ArcTanh[((a - b) Tan[x/2])/ Sqrt[-a^2 + b^2]])/((a^2 - b^2) Sqrt[-a^2 + b^2]) + (b Sin[x])/((-a^2 + b^2) (a + b Cos[x]))
所以
A = b/(b^2 - a^2);
B = (2 a)/(b^2 - a^2)

两边都取导对应的相等……